Derivative of 2^x: Solving Limits at 17

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In summary: So, to get the derivative of 2^x, we first write it as e^{xln(2)} and then apply the chain rule. In summary, To find the derivative of 2^x, we can use the chain rule after rewriting it as e^{xln(2)}. The derivative of e^x is simply e^x, and the constant C_a for a^x is equal to 1 when a is the number e, which is between 2 and 3.
  • #1
CrimeBeats
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Hi, I've been trying to find the derivative of 2^x and i got stuck here:
Lim (2^x - 1)/x
x=>0

How can i solve this limit (with steps please)
ps. I am only 17
 
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  • #2
CrimeBeats said:
Hi, I've been trying to find the derivative of 2^x and i got stuck here:
Lim (2^x - 1)/x
x=>0

How can i solve this limit (with steps please)
ps. I am only 17

Have you had the same limit for ex? What you want to do is notice that

2x = (eln(2))x = exln(2)

Now your limit becomes:

[tex]\lim_{x\rightarrow 0}\frac{e^{x\ln 2} - 1}{x}[/tex]

Now let u = x ln(2) and use what you know about ex.

[Edit] Corrected typo
 
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  • #3
I have been doing it this way all the time which I thought it is very straight forward:


[tex]y=2^{x}\Rightarrow ln(y)=xln(2)[/tex]

[tex]\Rightarrow \frac{1}{y}\frac{dy}{dx}=ln(2)[/tex]

[tex]\Rightarrow dy=d[2^{x}]=yln2dx=2^x ln(2)dx[/tex]
 
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  • #4
yungman said:
I have been doing it this way all the time which I thought it is very straight forward:


[tex]y=2^{x}\Rightarrow ln(y)=xln(2)[/tex]

[tex]\Rightarrow \frac{1}{y}\frac{dy}{dx}=ln(2)[/tex]

[tex]\Rightarrow dy=d[2^{x}]=xln(2)dx[/tex]
No, but probably a typo. You have to multiply both sides by y dx, not x dx.
[itex]dy= yln(2)dx= 2^x ln(2) dx[/itex].
 
  • #5
With [tex]e^{ln(2)x}[/tex], you can easily take the derivative of that. Just don't simplify the [tex]e^{ln2}[/tex] part. You should get [tex]ln(2)e^{ln(2)x}[/tex] as the numerator while the denominator becomes 1. Now do this expression with [tex]\lim_{x->0}[/tex]. By the way, LCKurtz made a slight typo.
 
  • #6
HallsofIvy said:
No, but probably a typo. You have to multiply both sides by y dx, not x dx.
[itex]dy= yln(2)dx= 2^x ln(2) dx[/itex].

My finger is get too far ahead of me! And I have destroy evidence already!:wink:
 
  • #7
I guess the answer is ln(2) * 2^x
No wonder i couldn't solve it before, they thaught us nothing about e

Depending on what you've said this means the derivative of e^x is e^x.. right?
 
  • #8
CrimeBeats said:
I guess the answer is ln(2) * 2^x
No wonder i couldn't solve it before, they thaught us nothing about e

Depending on what you've said this means the derivative of e^x is e^x.. right?

Yes. Most calculus books do that first before trying other bases.
 
  • #9
This problem might have been given in preparation for introducing "e".

The derivative of [itex]2^x[/itex] is
[tex]\lim_{h\to 0}\frac{2^{x+y}- 2^x}{h}= \lim_{h\to 0}\frac{2^x2^h- 2^x}{h}[/tex]
[tex]= \lim{h\to 0}\left(\frac{2^h- 1}{h}\right)2^x[/itex]
[tex]= \left(\lim_{h\to 0}\frac{2^h- 1}{h}\right)2^x[/tex]
That is, of course, a constant times [itex]2^x[/itex].

Similarly, the derivative of [itex]3^x[/itex] is
[tex]= \left(\lim_{h\to 0}\frac{3^h- 1}{h}\right)3^x[/tex]
a constant times [itex]3^x[/itex]

In that same way you can show that the derivative of [itex]a^x[/itex], for a any positive real number, is [itex]C_a a^x[/itex].
Further, by numerical approximations, you can show that [itex]C_2[/itex] is less than 1 and [itex]C_3[/itex] is greater than 1. There exists, then, a number, a, between 2 and 3 such that [itex]C_a= 1[/itex]. If we call that number "e", then the derivative of [itex]e^x[/itex] is just [itex]e^x[/itex] itself.
 

FAQ: Derivative of 2^x: Solving Limits at 17

What is the derivative of 2^x?

The derivative of 2^x is ln(2) * 2^x.

How do you solve limits involving 2^x?

To solve limits involving 2^x, you can use the rule that the limit of a^x as x approaches a is equal to a^a. In this case, a is equal to 2. So, the limit of 2^x as x approaches 2 is equal to 2^2, which is equal to 4.

What is the limit of 2^x as x approaches infinity?

The limit of 2^x as x approaches infinity is infinity. This is because as x gets larger and larger, 2^x also gets larger and larger, approaching infinity.

How do you find the derivative of 2^x at a specific point, such as x=17?

To find the derivative of 2^x at a specific point, such as x=17, you can use the power rule for derivatives. The power rule states that the derivative of x^n is equal to nx^(n-1). In this case, n is equal to 2, so the derivative of 2^x is 2x^(2-1) = 2x. Substituting x=17, we get a derivative of 34.

Can the derivative of 2^x be negative?

Yes, the derivative of 2^x can be negative. The derivative of 2^x is ln(2) * 2^x, and since ln(2) is a negative number, the derivative can be negative depending on the value of x.

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